我正在测试一个简单的网站。它在本地主机上运行,我可以在Web浏览器中访问它。索引页只是“运行”一词。urllib.urlopen
将成功读取页面,但urllib2.urlopen
将不成功。下面是一个演示问题的脚本(这是实际的脚本,不是对不同测试脚本的简化):
import urllib, urllib2
print urllib.urlopen("http://127.0.0.1").read() # prints "running"
print urllib2.urlopen("http://127.0.0.1").read() # throws an exception
这是堆栈跟踪:
Traceback (most recent call last):
File "urltest.py", line 5, in <module>
print urllib2.urlopen("http://127.0.0.1").read()
File "C:\Python25\lib\urllib2.py", line 121, in urlopen
return _opener.open(url, data)
File "C:\Python25\lib\urllib2.py", line 380, in open
response = meth(req, response)
File "C:\Python25\lib\urllib2.py", line 491, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python25\lib\urllib2.py", line 412, in error
result = self._call_chain(*args)
File "C:\Python25\lib\urllib2.py", line 353, in _call_chain
result = func(*args)
File "C:\Python25\lib\urllib2.py", line 575, in http_error_302
return self.parent.open(new)
File "C:\Python25\lib\urllib2.py", line 380, in open
response = meth(req, response)
File "C:\Python25\lib\urllib2.py", line 491, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python25\lib\urllib2.py", line 418, in error
return self._call_chain(*args)
File "C:\Python25\lib\urllib2.py", line 353, in _call_chain
result = func(*args)
File "C:\Python25\lib\urllib2.py", line 499, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 504: Gateway Timeout
有什么想法吗?我可能最终会需要一些更高级的功能,所以我不想仅仅依靠使用,而且我想了解这个问题。
最佳答案
听起来您已经定义了URLLIB2正在使用的代理设置。当它尝试代理“127.0.0.01/”时,代理放弃并返回504错误。
从Obscure python urllib2 proxy gotcha开始:
proxy_support = urllib2.ProxyHandler({})
opener = urllib2.build_opener(proxy_support)
print opener.open("http://127.0.0.1").read()
# Optional - makes this opener default for urlopen etc.
urllib2.install_opener(opener)
print urllib2.urlopen("http://127.0.0.1").read()